hdu 4033 二分几何
参考:http://blog.csdn.net/libin56842/article/details/26618129
题意:给一个正多边形内点到其他顶点的距离(逆时针给出),求正多边形的边长
二分多边形的边长
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" alt="" />
根据余弦定理求出A的角度,之后求出所有角度,加起来是否为360,小于则扩大,大于则缩小边
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define pi acos(-1.0)
#define exp 1e-8
int main()
{
int t,n,i,j,flag,cas = ;
double a[];
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
flag = ;
for(i = ; i<n; i++)
scanf("%lf",&a[i]);
a[n] = a[];
double l = ,r = ,mid;
for(i = ; i<=n; i++)
{
r = min(r,a[i]+a[i-]);
l = max(l,fabs(a[i]-a[i-]));
}
while(r-l>exp)
{
mid = (l+r)/;
double s,sum=;
for(i = ; i<=n; i++)
{
s = (a[i]*a[i]+a[i-]*a[i-]-mid*mid)/(2.0*a[i]*a[i-]);
sum+=acos(s);
}
if(fabs(sum-*pi)<exp)
{
flag = ;
break;
}
else if(sum>*pi)
r = mid;
else
l = mid;
}
printf("Case %d: ",cas++);
if(flag)
printf("%.3f\n",mid);
else
printf("impossible\n");
} return ;
}
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